Find parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1-t, z=2t and intersects this line.

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Novembruuo · Accepted Answer

Let us name (0,1,2) as the point A Point B is on the given line such that AB is perpendicular to the given line Vector along the given line r1=&amp;lt;1,−1,2&amp;gt; Vector along the given line A and B is r2=&amp;lt;1+t,1−t,2t≻&amp;lt;0,1,2≥&amp;lt;1+t,−t,2t−2&amp;gt; r1 and r2 must be perpendicular, hence their dot product must be 0 &amp;lt;1,−1,2&amp;gt;⋅&amp;lt;1+t,−t,2t−2≥0 (1+t)+t+(4t−4)=0 6t−3=0 6t=3 t=0.5 The required line passes through (0,1,2) Required line is parallel to r2=&amp;lt;1+0.5,−0.5,2⋅0.5−2≥&amp;lt;1.5,−0.5,−1&amp;gt; Recall that: Vector equation of a line passing a point with position vector a and parallel to the vector b is r(t)=a+tb Therefore equation of the required line is r(t)=&amp;lt;0,1,2&amp;gt;+t&amp;lt;1.5,−0.5,−1&amp;gt; r(t)=&amp;lt;1.5t,1−0.5t,2−t&amp;gt; Parametric equations are x=1.5t, y=1−0.5t, z=2−t Result: x=1.5t, y=1−0.5t, z=2−t

Vasquez · Accepted Answer

Step 1: Find the direction vector of the given line.The direction vector of the line x=1+t, y=1−t, z=2t is v→=(1,−1,2).Step 2: Find a vector perpendicular to v→.To find a vector perpendicular to v→, we can take the cross product of v→ with any other vector. Let&#039;s choose the vector u→=(1,0,0). The cross product of v→ and u→ is given by:n→=v→×u→=|𝐢𝐣𝐤1−12100|=(−2,−1,1)Step 3: Find the equation of the plane containing the given line and perpendicular to v→.The equation of a plane passing through the point (0,1,2) and with normal vector n→ is given by:−2x−y+z=dTo find the value of d, substitute the coordinates of the point (0,1,2) into the equation:−2(0)−(1)+(2)=dd=1Step 4: Find the intersection of the plane and the given line.Substitute the parametric equations of the line x=1+t, y=1−t, z=2t into the equation of the plane −2x−y+z=1 to find the intersection point:−2(1+t)−(1−t)+(2t)=1−2−2t−1+t+2t=1−3+t=1t=4Substituting t=4 back into the parametric equations of the line, we get the intersection point:x=1+4=5y=1−4=−3z=2(4)=8So, the intersection point is (5,−3,8).Step 5: Write the parametric equations for the line.We can write the parametric equations for the line as follows:x=5ty=−3t+1z=8t+2Therefore, the parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1−t, z=2t and intersects this line are x=5t, y=−3t+1, and z=8t+2.

user_27qwe · Accepted Answer

To find the parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1−t, z=2t and intersects this line, we can follow these steps:Step 1: Determine the direction vector of the given line.The given line is described by the equations x=1+t, y=1−t, z=2t. We can observe that the direction vector of this line is d→=⟨1,−1,2⟩.Step 2: Find the direction vector of the line perpendicular to the given line.Since we want a line that is perpendicular to the given line, we can take the dot product of the direction vectors of the two lines and set it equal to zero. This will ensure that the lines are orthogonal.Using the dot product formula, we have:d→·n→=0where n→=⟨a,b,c⟩ is the direction vector of the line we are trying to find.Substituting the values of d→ and n→, we get:⟨1,−1,2⟩·⟨a,b,c⟩=0This simplifies to:a−b+2c=0Step 3: Find the parametric equations of the line.To find the parametric equations, we need a point on the line. We are given the point (0,1,2), which lies on the line we are trying to find. Let&#039;s call this point (x0,y0,z0). Therefore, x0=0, y0=1, and z0=2.The parametric equations for the line are given by:x=x0+aty=y0+btz=z0+ctSubstituting the values of x0=0, y0=1, and z0=2, we have:x=0+aty=1+btz=2+ctFinally, we substitute the value of b in terms of a and c using the equation a−b+2c=0. Solving for b, we get b=a+2c. Substituting this into the parametric equations, we have:x=0+aty=1+(a+2c)tz=2+ctTherefore, the parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1−t, z=2t and intersects this line are:x=ty=1+(t+2c)tz=2+ctwhere a and c are any real numbers.

karton · Accepted Answer

Answer:x=0y=1+2sz=2+sExplanation:To find the parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1−t, z=2t and intersects this line, we can follow these steps:1. Determine the direction vector of the given line. From the given parametric equations, we can see that the direction vector of the line is 𝐯=⟨1,−1,2⟩.2. To find a vector perpendicular to 𝐯, we can take the cross product of 𝐯 with any non-parallel vector. Let&#039;s choose 𝐰=⟨1,0,0⟩. The cross product is given by:𝐧=𝐯×𝐰=|𝐢𝐣𝐤1−12100|Evaluating the determinant, we get:𝐧=⟨0,2,1⟩3. Now we have the direction vector 𝐧 of the line we seek, and we also have a point (0,1,2) on the line. We can use this information to write the parametric equations of the line.Let 𝐫=⟨x,y,z⟩ be a point on the line. We can write:𝐫=⟨0,1,2⟩+s⟨0,2,1⟩where s is a parameter representing the distance along the line. This gives us the desired parametric equations for the line:x=0+0sy=1+2sz=2+sTherefore, the parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1−t, z=2t, and intersects this line are:x=0y=1+2sz=2+s

Find parametric equations for the line through the point (0,1,2)

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